Use the work–energy theorem to solve each of these problems. You can use Newton’

Question

Use the work–energy theorem to solve each of these problems. You can use Newton’s laws to check your answers.

a) A skier moving at 4.87 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping?

b) Suppose the rough patch in part A was only 2.85 m long. How fast would the skier be moving when she reached the end of the patch?

c) At the base of a frictionless icy hill that rises at 25.0 above the horizontal, a toboggan has a speed of 12.9 m/s toward the hill. How high vertically above the base will it go before stopping?

Explanation / Answer

Ok, friction depends upon the normal reaction of the ground/earth on the skier. Assuming the plane to be perfectly horizontal with no inclination, the total reaction equals the total weight, W. Let the coefficient be u - so friction force acting becomes uW.

Assuming no forward force/thrust, the only force in the direction of motion is friction.
Let the mass of skier be m.

As W=mg where g is acceleration due to gravity, friction f = u(m x g)
Net acceleration, a = f / m (Second Law)

a = - u(m x g) / m = - u x g = -0.22 * 9.81 = -2.15 m/s^2


Using equation of motion, v^2 - u^2 = 2as,
a = -2.15
v = 0 (when stopped)
u = 4.87

0-4,87^2 = 2(-2.15)(s)

so, s =
s from calculation = 5.51m

b) For 2.85m patch,
a = -2.15
u = 4.87
s = 2.85

v^2-4.87^2 = 2(-2.15)(2.85)
v from calculation = 5.99 m/s

c) KE at foot of hill

= 0.5 * m * 12.9 ^2

PE when it stops

= m * g * h

KE =PE

0.5 * m * 12.9 ^2 = mgh

0.5 * 12.9 ^2 = 9.81 * h

h = 8.48 m above the base will it go before stopping

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